Taken from wolfram:
It is probably something very simple, but I can't quite get where the factor of $\frac{1}{2}$ comes from.
Given $z = x + iy,$ I would think that (and maybe this is where I'm going wrong)
$ \frac{\partial x}{\partial z} = 1,$ and
$ \frac{\partial y}{\partial z} = -i,$
therefore (4) can be expressed as
$ \frac{df}{dz} = \frac{\partial f}{\partial x} (1) + \frac{\partial f}{\partial y} (-i)$.
The only way the math would work is if $\frac{\partial x}{\partial z} = \frac{1}{2}$ and $\frac{\partial y}{\partial z} = -\frac{i}{2}$, but I can't see how to show that formally.
(5) is certainly correct when I plug in specific examples.
Hint: \begin{align} x = \frac{z+\bar z}{2} \ \ \text{ and } \ \ y =\frac{z-\bar z}{2i} \end{align} which means \begin{align} \frac{\partial x}{\partial z} = \frac{1}{2} \ \ \text{ and } \ \ \frac{\partial y}{\partial z} = \frac{1}{2i}. \end{align}